If the range of function ` f(x) = (x + 1)/(k+x^(2))` contains the interval
[-0,1] , then values of k can be equal to

Options

A. 0
B. 0.5
C. 1.25
D. 1.5

Explanation

To solve the problem, we need to determine the values of \( k \) such that the range of the function

\[ f(x) = \frac{x + 1}{k + x^2} \]

contains the interval \([-0, 1]\).

### Step 1: Rewrite the function

We start by rewriting the function in a form that allows us to analyze its behavior. We can set \( y = f(x) \):

\[ y = \frac{x + 1}{k + x^2} \]

### Step 2: Rearranging the equation

Rearranging the equation gives:

\[ y(k + x^2) = x + 1 \]

This can be rewritten as:

\[ yx^2 - x + (yk - 1) = 0 \]

### Step 3: Identify coefficients

This is a quadratic equation in \( x \) of the form \( ax^2 + bx + c = 0 \), where:

- \( a = y \)
- \( b = -1 \)
- \( c = yk - 1 \)

### Step 4: Discriminant condition

For the quadratic equation to have real roots, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Substituting the coefficients:

\[ (-1)^2 - 4(y)(yk - 1) \geq 0 \]

This simplifies to:

\[ 1 - 4y(yk - 1) \geq 0 \]

### Step 5: Expand and rearrange the inequality

Expanding this gives:

\[ 1 - 4y^2k + 4y \geq 0 \]

Rearranging leads to:

\[ 4ky^2 - 4y - 1 \leq 0 \]

### Step 6: Analyze the quadratic inequality

This is a quadratic inequality in \( y \). For the quadratic \( 4ky^2 - 4y - 1 \) to be less than or equal to zero, we need to analyze its roots.

### Step 7: Find the roots

Using the quadratic formula, the roots are given by:

\[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4k)(-1)}}{2(4k)} = \frac{4 \pm \sqrt{16 + 16k}}{8k} \]

This simplifies to:

\[ y = \frac{1 \pm \sqrt{1 + k}}{2k} \]

### Step 8: Determine the range of \( k \)

For the range of \( y \) to contain the interval \([-0, 1]\), we need to evaluate the values of \( k \) such that:

1. The maximum root \( \frac{1 + \sqrt{1 + k}}{2k} \geq 1 \)
2. The minimum root \( \frac{1 - \sqrt{1 + k}}{2k} \leq 0 \)

### Step 9: Solve the inequalities

1. For the maximum root:

\[ \frac{1 + \sqrt{1 + k}}{2k} \geq 1 \implies 1 + \sqrt{1 + k} \geq 2k \implies \sqrt{1 + k} \geq 2k - 1 \]

Squaring both sides gives:

\[ 1 + k \geq 4k^2 - 4k + 1 \implies 4k^2 - 5k \leq 0 \implies k(4k - 5) \leq 0 \]

This gives \( 0 \leq k \leq \frac{5}{4} \).

2. For the minimum root:

\[ \frac{1 - \sqrt{1 + k}}{2k} \leq 0 \implies 1 - \sqrt{1 + k} \leq 0 \implies \sqrt{1 + k} \geq 1 \implies 1 + k \geq 1 \implies k \geq 0 \]

### Final Step: Combine the results

Combining the results from both inequalities, we find:

\[ 0 \leq k \leq \frac{5}{4} \]

Thus, the values of \( k \) can be:

\[ k \in [0, \frac{5}{4}] \]